\(2x-2y+4xy=6\Leftrightarrow2y\left(2x-1\right)+2x-1=5\)
\(\Leftrightarrow\left(2y+1\right)\left(2x-1\right)=5\)
TH1: \(\left\{{}\begin{matrix}2y+1=5\\2x-1=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}y=2\\x=1\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}2y+1=1\\2x-1=5\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}y=0\\x=3\end{matrix}\right.\)
TH3: \(\left\{{}\begin{matrix}2y+1=-1\\2x-1=-5\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}y=-1\\x=-2\end{matrix}\right.\)
TH4: \(\left\{{}\begin{matrix}2y+1=-5\\2x-1=-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}y=-3\\x=0\end{matrix}\right.\)
Ta có : \(x-y+2xy=3\)
\(\Leftrightarrow\) \(2x-2y+4xy=6\)
\(\Leftrightarrow\) \(2x+4xy-\left(2y+1\right)=5\)
\(\Leftrightarrow\)\(2x\left(1+2y\right)\)\(-\left(1+2y\right)\)\(=5\)
\(\Leftrightarrow\)\(\left(2x-1\right)\left(2y+1\right)=5\)
Đến đây ta kẻ bảng xét các giá trị của \(x,y\).
Các cặp số nguyên \(\left(x,y\right)\)thoả mãn bài toán là:
\(\left(x,y\right)\)∈\(\left\{\left(3,1\right);\left(1,3\right);\left(-2,0\right);\left(0,-2\right)\right\}\)
Vậy \(\left(x,y\right)\)∈\(\left\{\left(3,1\right);\left(1,3\right);\left(-2,0\right);\left(0,-2\right)\right\}\)
